Theo vi-et, ta có:
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=4\\x_1x_2=\dfrac{c}{a}=-\dfrac{5}{3}\end{matrix}\right.\)
\(T=\dfrac{x_1^2+4x_2-x_1x_2}{4x_1+x_2^2+x_1x_2}\)
\(=\dfrac{x_1^2+x_2\left(x_1+x_2\right)-x_1x_2}{x_1\left(x_1+x_2\right)+x_2^2+x_1x_2}=\dfrac{\left(x_1^2+x_2^2\right)}{\left(x_1+x_2\right)^2}\)
\(=\dfrac{\left(x_1+x_2\right)^2-2x_1x_2}{\left(x_1+x_2\right)^2}=\dfrac{4^2-2\cdot\dfrac{-5}{3}}{4^2}=\dfrac{29}{24}\)